14 comments:

More - take it to either extreme - if he's level with the boat it will advance one yard for one yard of string. If he's almost above it, then the difference in rope-length between the boat being a few feet from shore and aground will be next to nothing.

Unfortunately, the answer and explanation at the link is wrong. Sure, b > a-c, but that equation does not determine the absolute relationship of b and a. If you consider the starting equation of b+c>a then you notice that you must add something to b in order to exceed a -- so a must be the greater number. (To be precise, these equations should be >= for the case of c=0.)

The correct answer is less (or equal if c=0). Consider the extreme case where the boy is on the shore (c=0), then the rope and the distance traveled are the same (a=b). If you stand above the shore (c>0), then it takes a longer rope to reach the object in the water. Thus, when you pull in 1m of rope, the boat will move some smaller distance.

As an English major, I don't mind sounding stupid re math, so I'll ask what may be a dumb question. I don't see any need for the equation to "determine the absolute relationship of b and a." It doesn't seem to matter if the ratio is 1:1.5 or 1:10.

And I agree that when the puller is up high, it "takes a longer rope to reach the object", but it doesn't logically follow that "when you pull in 1m of rope, the boat will move some smaller distance. Why can't it move more than a yard?

I too have a problem with this. First, the problem remains the same if the "C" side is at right angles to the shore line, making this a a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.

In the original condition, we have b*b = a*a - c*c, and in the second case, we have b*b = (a-1)*(a-1) - b'*b'.

Argh. I changed some variable names from my paper calculation, and messed up the post. Let me try again!

I too have a problem with this. First, the problem remains the same if the "c" side is at right angles to the shore line, making this a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.

In the original condition, we have c*c = a*a - b*b, and in the second case, we have c*c = (a-1)*(a-1) - b'*b'.

One more time. Need to get more than 2 hours of sleep a night. I mashed a sign near the end:

Argh. I changed some variable names from my paper calculation, and messed up the post. Let me try again!

I too have a problem with this. First, the problem remains the same if the "c" side is at right angles to the shore line, making this a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.

In the original condition, we have c*c = a*a - b*b, and in the second case, we have c*c = (a-1)*(a-1) - b'*b'.

I've finally figured out why this problem's solution is so counterintuitive. When we hear, "pull in one unit on the hypotenuse," the thought immediately is "similar triangles!" in which case the boat will approach the shore at less than one unit.

BUT: These are NOT similar triangles; the angle at the point of pull, relative to vertical, becomes smaller. The boat is then much like a pendulum approaching its own vertical.

Just from fishing experience and not bothering with math, the bobber reels in faster as it gets closer to the rod. It's faster still when you're on an elevated pier. So the solution is pretty intuitive for me.

(A) One yard pulled in equals one yard of boat distance reduction.Unless you have some rope that elastically stretches out when pulled or magically shrinks when pulled.

This isn't a trigonometry problem. It is basic, pulling rope 1 foot results in object at the end of the rope getting nearer to you by 1 foot problem.

Here is a similar question with the above diagram.The boy is holding a long thin semi-rigid metal rod. The end of the rod is 10 feet away. The boy while holding the rod steps 1 foot backwards, will the end of the rod also move 1 foot closer to the shore?This isn't a trigonometry problem, it is a low-grade intelligence test.

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